While a capacitor is connected to a battery, the plate , separation is halved, f
ID: 2003702 • Letter: W
Question
While a capacitor is connected to a battery, the plate , separation is halved, from d to d/2. State battery how this affects the following. Express answers in terms of the original value. The capacitor starts with no dielectric material between the plates. The plates are brought back to their original separation d with no dielectric between them, and then the battery is disconnected. Then, a dielectric of constant kappa = 3 is inserted, filling the space between the plates. State how this affects the following. Express answers in terms of the original value. With the plates at their original separation d with no dielectric between them, the battery is re-connected. Then, the area of the plates, A, is increased by a factor of 3. The battery remains connected. State how changing the area affects the following. Express answers in terms of the original value.Explanation / Answer
a) The applicable electrical relations are:
C0 = A/d, Q0 = C0V0, V0 = Q0/C0, E0 = V0/d , U0 = ½C0V0²
if d' = d/2
then C' = A/d' = 2A/d = 2C0, => capacitance double.
Q' = C'V0 = 2C0V0, then Q' = 2Q0 => charge double
V potential difference unchanged.
E' = V0/d' = 2V0/d = 2E0, => electric field double.
U' =0.5C'V02 =0.5*2C0V02 =2U0 => energy is doubled.
b) Once, the charged capacitor is disconnected from the battery, and in the absence of leakage current, the
voltage remains constant. Dielectric constant k=3
V' = V0/k = V0/3
E' = V'/d = V0/3d = E0/3,
C' =kC0 =3C0
U' =0.5C'V'2 =0.5*3C0V02/9 =U0/3
c) With A'= 3A then C' = A'/d = 3C0
Q' =C'V0 = 3Q0
E' = V0/d =E0
U' = 0.5C'V0^2 =3U0
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